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lara [203]
3 years ago
8

A string is wrapped several times around the rim of a small hoop with radius 7cm and mass 2kg. The free end of the string is hel

d in place and the hoop is released from rest.
After the hoop has descended 80cm, calculate (a) the angular speed of the hoop and (b) the speed of its center.
Physics
1 answer:
Ivanshal [37]3 years ago
3 0

Answer:

Explanation:

Given that,

The radius of loop r = 7cm = 0.07m

Mass of hoop M = 2kg

The hoop is released from rest, the initial velocity is 0m/s

A. Angular speed of the hoop after a descended of h= 80cm = 0.8m

Applying the conservation of energy

Ei + W = Ef

Where,

Ei = initial energy of the system, I.e the initial potential and kinetic energy

Ef = final energy of the system I.e the final potential and kinetic energy

W = 0, since no external force is acting on the systems

Ui + K.Ei + 0 = Uf + K.Ef

System was initially at rest, K.Ei = 0

Uf = 0 at the zero level

Then,

Mgh = ½ •Icm•w² + ½M•Vcm²

Icm = Mr², since it is circular loop

Then,

Mgh = ½ •Mr²•w² + ½M•Vcm²

M cancels out

gh = ½ •r²•w² + ½ Vcm²

Since Vcm = wr

Then, gh = ½ •r²•w² + ½ w²r²

gh = r²w²

w = √(gh/r²)

w = √(9.81 × 0.8/0.07²)

w = 40 rad/s

b. Speed at the center

Since Vcm = wr

Vcm = 40×0.07

Vcm = 2.8 m/s

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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
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Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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